Given a list of letters,
letters = { "A", "B", ..., "F" }
is it possible to get Mathematica to generate all ‘words’ (in this example, 6 letter words), if only one letter can be used one time only, e.g. ABCDEF, ABCDFE, …? TIA.
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Pemutations will do it:
letters = {"a", "b", "c"};
Permutations[letters, {3}]
{{"a", "b", "c"}, {"a", "c", "b"}, {"b", "a", "c"},
{"b", "c", "a"}, {"c", "a", "b"}, {"c", "b", "a"}}
To get all six-letter words:
letters = {"a", "b", "c", "d", "e", "f"};
perms = Permutations[letters, {6}];
StringJoin /@ perms
{"abcdef", "abcdfe", "abcedf", "abcefd", "abcfde" ... etc.
there are a lot of them.
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You can create permutations with all of the letters as strings with:
StringJoin /@ Permutations[letters]
If you want lists of the individual letters just use:
Permutations[letters]
Check the documentation of Permutations to learn about permutations with subsets of letters. If you want to use each letter more than once, look at the documentation for Tuples.
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$\begingroup$ Thanks(x2). Is there some way to check how many words contain a ‘sub-word’, like ‘ab’ or even a set of ‘sub-words’ like ‘ab’ and ‘cd’? And is there any web page or text book that deals with combinatorics in Mathematica (on a more ‘basic’ level) that I could visit/buy and read? $\endgroup$ – mf67 Mar 24 '19 at 2:54
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If I follow the OP's question, I think they want the following:
letters = {"a", "b", "c"};
p = Permutations[letters, {#}] & /@ Range[Length[letters]];
(StringJoin[#] & /@ #) & /@ p
{{a, b, c}, {ab, ac, ba, bc, ca, cb}, {abc, acb, bac, bca, cab, cba}}
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$\begingroup$ No, the OP requested all six letter words. $\endgroup$ – m_goldberg Mar 24 '19 at 11:22
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$\begingroup$ Your last line can be written more cleanly as
Map@StringJoin/@porMap[StringJoin, p, {2}]. $\endgroup$ – Doorknob Mar 24 '19 at 22:48
